python - Why is the global keyword not required in this case?

Question

cache = {}
def func():
    cache['foo'] = 'bar'
print cache['foo']

output

bar

Why does this work and why doesn't it require use of the global keyword?

@AshwiniChaudhary That's simply untrue. Try this with a list and it won't work and lists and mutable! mylist = [] def f1(): mylist = ['a']; mylist is still empty if you print it outside of f1() — samsamara
@samsamara That's not what I meant, with mylist = ['a'] you're explicitly defining a new local variable. Perform a mutation instead, like mylist.append(10) etc — Ashwini Chaudhary
Yes if you do a mutation like that, it is going to work as you are only accessing it. But say if i change my function to accept a list as a parameter, def f1(mylist): mylist = mylist + ['a']; where I'm mutating the list using the assignment operator. And I have mylist = ['s'] and I call f1(mylist). Now as lists are mutable and you passing list object to f1() yet it is still not mutating mylist. — samsamara

Martijn Pieters Martijn Pieters · Accepted Answer · 2012-12-29T11:46:17

Because you are not assigning to cache, you are changing the dictionary itself instead. cache is still pointing to the dictionary, thus is itself unchanged. The line cache['foo'] = 'bar' translates to cache.__setitem__('foo', 'bar'). In other words, the value of cache is a python dict, and that value is itself mutable.

If you tried to change what cache refers to by using cache = 'bar' instead, you would be changing what cache points to and then you need the global keyword.

Perhaps this older answer of mine to a similar question helps you understand the difference: Python list doesn't reflect variable change.

python - Why is the global keyword not required in this case?

1 Answers